(Since this question is so old, I suspect it is safe to provide an answer.)

I think you're halfway there. We might often use a $t$-test in a situation like this, but a confidence interval is not quite the same thing as a test. Since the mean and SD are estimated from the data, you need to take that fact into account. Thus, we will use the $t$-distribution to form the confidence interval. The general formula would be:
$$ \bar x \pm t_{1-\frac{\alpha}{2},\ df}\ \frac{s}{\sqrt{N}} $$ The key to using this formula is to find the relevant $t$-value. First, we need to get the df—it is $N-1=54$. Then we look up the quantile that corresponds to the $99.5^{\rm th}$ percentile of that particular $t$-distribution in a $t$-table. I find the value $2.67$. Hence, $$ 10.5 \pm 2.67\ \frac{3.25}{\sqrt{55}} = 10.5 \pm 1.17 \Rightarrow (9.33,\ 11.67) $$

(Since this question is so old, I suspect it is safe to provide an answer.)

I think you're halfway there. We might often use a $t$-test in a situation like this, but a confidence interval is not quite the same thing as a test. Since the mean and SD are estimated from the data, you need to take that fact into account. Thus, we will use the $t$-distribution to form the confidence interval. The general formula would be:
$$ \bar x \pm t_{(1-\frac{\alpha}{2}\!,\ df)}\ \frac{s}{\sqrt{N}} $$ The key to using this formula is to find the relevant $t$-value. First, we need to get the df—it is $N-1=54$. Then we look up the quantile that corresponds to the $99.5^{\rm th}$ percentile of that particular $t$-distribution in a $t$-table. I find the value $2.67$. Hence, $$ 10.5 \pm 2.67\ \frac{3.25}{\sqrt{55}} = 10.5 \pm 1.17 \Rightarrow (9.33,\ 11.67) $$